Sampling frequency or Nyquist rate; f_{s} = 2 × 150 = 300 Hz

2. Consider two real valued signals, x(t) band-limited to [- 500 Hz, 500 Hz] and y(t) band-limited to [-1 kHz, 1 kHz]. For z(t) = x(t)⋅y(t), the Nyquist sampling frequency (in kHz) is ______.

3 kHz

5 kHz

10 kHz

15 kHz

Answer.1. 3 kHz

Explanation

Convolution theorem states that the multiplication in time domain is equal to convolution in frequency domain, and vice-versa, i.e.

z(t) = x(t) ⋅ y(t) ⇔ Z(f) = X(f) * Y(f)

where X(f) and Y(f) are the Fourier transform of x(t) and y(t) respectively.

The Nyquist sampling rate is given as:

f_{s} = 2 × f_{max}

f_{max} = Maximum frequency present at the input

Application:

Given x(t) is band-limited to [-500 Hz, 500 Hz]

y(t) is band-limited to [-1 kHz, 1 kHz]

z(t) = x(t)⋅y(t)

Since Z(f) = X(f) * Y(f), the output spectrum will be:

∴ The range of convolution in the frequency domain is [-1500 Hz, 1500 Hz]

So, the maximum frequency present in z(t) will be 1500 Hz.

The Nyquist rate will now be:

f_{s} = (2 f_{max}) is 3000 Hz or 3 kHz

3. The spectrum of a bandpass signal spans from 20 kHz to 30 kHz. The signal can be recovered ideally from the sampled values when the sampling rate is at least:

20 kHz

60 kHz

50 kHz

40 kHz

Answer.1. 20 kHz

Explanation

Sampling frequency of bandpass signal is given by:

4. If the sampling is carried out at a rate higher than twice the highest frequency of the original single (fmax), then it is possible to receive the original signal from the sampled signal by passing it through

A high-pass filter with the cut-off frequency equal to f_{max}

A low-pass filter with the cut-off frequency equal to f_{max}

A high-pass filter with the cut-off frequency greater than f_{max }

A low-pass filter with the cut-off frequency lesser than f_{max }

Answer.2. A low-pass filter with the cut-off frequency equal to f_{max}

Explanation

In the sampling process, if the sampling frequency is greater than the Nyquist frequency for the reconstruction of the original signal, LPF is used.

The sampling frequency can take three possible values with respect to spectrum width (2 ω_{m})

(i) ω_{s} > 2ω_{m}

(ii) ω_{s} = 2 ω_{m}

(iii) ω_{s} < 2ω_{m}

If the sampling is carried out at rate higher than twice the highest frequency of the original single (fmax), then it is possible to receive the original signal from the sampled signal by passing it through A low-pass filter with the cut-off frequency equal to f_{max}

When ω_{s} = 2ω_{m}

The spectrum of the sampled signal will be:

We see for ω_{s} > 2ω_{m}, there is no overlap between the shifted spectrum of X(ω). Thus, as long as the sampling frequency ω_{s} is greater than the twice the signal bandwidth (2 ω_{m}), x(t) can be recovered by passing the sampled signal x_{s}(t) through an ideal or practical low pass filter having bandwidth between ω_{m} and (ω_{s} – ω_{m}) rad/sec.

5. A system has a sampling rate of 50,000 samples per second. The maximum frequency of the signal it can acquire to reconstruct is

25 kHz

50 kHz

100 kHz

10 kHz

Answer.1. 25 kHz

Explanation

Nyquist Sampling Theorem:

A continuous-time signal can be represented in its samples and can be recovered back when sampling frequency fs is greater than or equal to twice the highest frequency component of the message signal, i.e.

fs ≥ 2fm

Therefore when we want to convert continuous signals to discrete signals, the sampling must be done at the Nyquist rate.

Application:

Given f_{s} = 50 k

We can write:

50k ≥ 2 × f_{m}

f_{m} ≤ 25 k

∴ The maximum frequency of the signal it can acquire to reconstruct is 25 kHz

6. A band-limited signal is sampled at Nyquist rate, The signal can be recovered by passing the samples through

RC filter

envelope detector

PLL

ideal low pass filter with appropriate bandwidth

Answer.4. ideal low pass filter with appropriate bandwidth

Explanation

Nyquist Sampling Theorem:

A continuous-time signal can be represented in its samples and can be recovered back when sampling frequency fs is greater than or equal to twice the highest frequency component of the message signal, i.e.

fs ≥ 2fm

Therefore when we want to convert continuous signals to discrete signals, the sampling must be done at the Nyquist rate.

If the low pass filter is not ideal, just sampling at the Nyquist frequency will not work we need to sample above the Nyquist frequency.

For example:

We have assumed spectrum x(t) is band limited to ωm.

Let, ωs > 2ωm

Let ωs = 3ωm

The spectrum of the sampled signal will be:

We see for ωs > 2ωm, there is no overlap between the shifted spectrum of X(ω). Thus, as long as the sampling frequency ωs is greater than the twice the signal bandwidth (2 ωm), x(t) can be recovered by passing the sampled signal xs(t) through an ideal or practical low pass filter having bandwidth between ωm and (ωs – ωm) rad/sec.

7. If the lower sideband overlaps the basebands, the distortion is called

Cross-Over distortion

Aliasing

Crosstalk

None of these

Answer.2. Aliasing

Explanation

If the lower sideband overlaps the basebands, the distortion is called Aliasing. The main reason for aliasing is undersampling i.e fs < 2fm.

8. What is the maximum bit rate of a noiseless channel with a bandwidth of 1000 Hz transmitting a signal with two signal levels?

2000 bps

3000 bps

4000 bps

6000 bps

Answer.1. 2000 bps

Explanation

Given,

Bandwidth = 1000 Hz

L = 2

Hence, Maximum bit rate = 2 × bandwidth × log2 L = 2 × 1000 × log_{2}2 = 2000 bps

9. Consider a real-value based-band signal x(t), band limited to 10 kHz. The Nyquist rate for the signal y(t) = x(t) $x(1 + \frac{t}{2})$ is

$y\left( t \right) = x\left( t \right) \cdot x\left( {\frac{t}{2} + 1} \right)$

Nyquist rate of y(t) = 2 [frequency of y(t)]

= 2 × 15 kHz

= 30 kHz

10. A sinusoid of 10 kHz is sampled at 15 k samples/s. The resulting signal is passed through an ideal low pass filter (LPF) with a cut-off frequency of 25 kHz. The maximum frequency component at the output of the LPF (in kHz) is

25 kHz

20 kHz

10 kHz

40 kHz

Answer.1. 25 kHz

Explanation

After sampling, frequencies present will be: ± f_{m} ± nf_{s}

Where, n = 0, 1, 2, 3, 4, …

For n = 0, frequencies present = ±10 kHz

For n = 1, frequencies present = ± 10 k ± 15 k = ±25 kHz, ±5 kHz

After passing through the ideal LPF having a cut-off frequency of 25 kHz, the highest frequency component present at filter output = 25 kHz.

Therefore, the maximum frequency present at filter output is 25 kHz